A person walks 1.6 km north. Then he changes direction and walks 2.5km at 25 degrees north of west. What is the total displacement from the original point?

A. 4.1 km at 25º north of west
B. 3.1 km at 43º west of north
C. 3.5 km at 50º north of west
D. 4.1 km at 47º west of north
E. 4.1 km at 65º north of west

Question

A person walks 1.6 km north.  Then he changes direction and walks 2.5km at 25 degrees north of west.  What is the total displacement from the original point?

A.	4.1 km at 25º north of west
B.	3.1 km at 43º west of north
C.	3.5 km at 50º north of west
D.	4.1 km at 47º west of north
E.	4.1 km at 65º north of west

anonymous · Answer

it is a vector

anonymous · Answer

The square root of the sum of the distances squared

anonymous · Answer

You have to draw it out. There's no movement in a negative direction (east) so you just add the displacement like distance. 1.6+2.5 = 4.1 km. Also since traveling north, and only turned 25 degrees north west.. your direction stays 25 degrees north of west.

anonymous · Answer

Well I thought that North of West would make it 25 degrees away from West?  Or am I wrong about this?

anonymous · Answer

|dw:1349731676296:dw|
$$d=\sqrt{a^2+b^2+2ab\cos\theta}\\tan\alpha=\frac{y}{x}$$

anonymous · Answer

So would the answer be A?

anonymous · Answer

Also what is the second part of the formulas? Tan theta=y/x?

anonymous · Answer

The second part is used to find direction theta..

anonymous · Answer

So you would set it up as 1.6/2.5=.64 and it would be 64~65 degrees?